Fun Math Challenge: Can You Find the Diameter?
14K views · Sep 25, 2026 · Education
Comments · 7
@RamanaCv-o2l · 2 days ago
For this type of same problems, add the diameters of two semi circles. It gives diameter of required semi circle. So it is 5.
@crimsonknight5575 · 5 days ago
Tough one
1
@MichaelPaoli · 22 hours ago
Draw a triangle with sides from<br>radial center of diameter 2 semicircle to<br>radial center of diameter 3 semicircle,<br>then straight down from there to<br>bottom of diameter 3 semicircle, and then from there back to<br>radial center of diameter 2 semicircle, forming right triangle.<br>Hypotenuse goes straight through where the 2 smaller semicircles are<br>tangent, thus it has length the sum of their radii = 2/2+3/2=<br>2.5.<br>Right side has length 3/2=1.5<br>We can get the remaining side from Pythagorean Theorem.<br>xx+1.5^2=2.5^2<br>xx=2.5^2-1.5^2<br>x=(2.5^2-1.5^2)^.5<br>x=2<br>We now have what we can determine formula for our largest (semi)circle,<br>and thus radius and diameter. It's general formula, we place it radial<br>center on the x axis, thus have:<br>(x+a)^2+yy=rr, y>=0<br>Let's place leftmost point of largest semicircle at (0,0), so we have:<br>(x-r)^2+yy=rr, x>=0, y>=0, r>0<br>We have additional point on circle, where the two larger semicircles<br>touch above the x axis. Given the Cartesian orientation and placement<br>chosen, for that point, we have x=2/2+2+3/2=4.5, y=3/2=1.5<br>(4.5-r)^2+(1.5)^2=rr, x>=0, y>=0, r>0 <br>20.25-9r+rr+2.25=rr<br>9r=22.5=45/2<br>r=5/2<br>d=2r=2*5/2<br>d=5
@משהנאון-י6מ · 3 days ago
X=5.05
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